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The four numbered areas are congruent. [Added later] The figure below is from a suggested edit by @TomZych, and it shows the congruent parts more clearly. Given all the upvotes to the (probably tongue-in-cheek) comment “This answer also deserves the tick for artistic reasons,” I’m leaving my original “artistic” figure but also adding Tom’s improved version ...


It is probably easier to note that $2^{40} = 4^{20}$. The only ones of the 20 factors in $20!$ that are smaller than $4$ are $1$, $2$ and $3$. But, on the other hand, $18$, $19$ and $20$ are all larger than $4^2$, so we can see $$ 20! = 1\cdot 2\cdot3\cdots 18\cdot19\cdot 20 > \underbrace{1\cdot1\cdot1}_{3\text{ ones}}\cdot\underbrace{4\cdot4\cdots 4\cdot ...


You have to be careful when multiplying by $x$ since $x$ might be negative and hence flip the inequality. Suppose $x>0$. Then $$\frac{1}{x}<4\iff4x>1\iff x>1/4.$$ If $x>0$ and $x>1/4$, then $x>1/4$. Now suppose $x<0$. Then $$\frac{1}{x}<4\iff4x<1\iff x<1/4.$$ If $x<0$ and $x<1/4$, then $x<0$. So the solution set is ...


I think sketching the two identical triangles marked with green below makes this rather intuitive. This could also be turned into a formal proof quite easily.


Here is the solution $$\frac { 1 }{ x } <4$$$$ \frac { 1-4x }{ x } <0$$$$ \frac { x\left( 1-4x \right) }{ { x }^{ 2 } } <0$$$$ x\left( 1-4x \right) <0$$$$ x\left( 4x-1 \right) >0 $$ so $$x\in \left( -\infty ,0 \right) \cup \left( \frac { 1 }{ 4 } ,+\infty \right) $$


Another way (not sure if its "simple" though!): $y = x+1$ is the tangent line to $y = e^x$ when $x= 0$. Since $e^x$ is convex, it always remains above its tangent lines.


I suggest $$\sum_{k=1}^n \frac{1}{k^2} \leqslant 1 + \sum_{k=2}^n \frac{1}{k^2 - \frac14} = 1 + \sum_{k=2}^n \left(\frac{1}{k-\frac12} - \frac{1}{k+\frac12}\right) = 1 + \frac23 - \frac{1}{n+\frac12}.$$




Inscribe a regular hexagon in a circle of radius $1$. Since a straight line is the shortest distance between two points the circumference of the circle is longer than the circumference of the hexagon. We take the definition of $\pi$ as half the circumference of the unit circle. Putting all this together we obtain $2\pi \gt 6$ or $\pi \gt 3$ We take $e$ as ...


Side note: it's actually the Cauchy-Schwarz-Bunyakovsky inequality, and don't let anyone tell you otherwise. The problem with using the geometric definition is that you have to define what an angle is. Sure, in three dimensional space, you have pretty clear ideas about what an angle is, but what do you take as $\theta$ in your equation when $i$ and $j$ are $...


Your inequality is equivalent to $$(7!)^8 < (8!)^7$$ divide it by $(7!)^7$, and get $$7! < 8^7$$ and this is clear, since $$1 \cdots 7 < 8 \cdots 8$$


You can use calculus to find where the maximum of $f(x)=x^{1/x}$ occurs. You will find that it is at $x=e$. You will also find that $f(x)$ is increasing for $0< x<e$ and decreasing for $e<x$. Unfortunately, the choice $e^{1/e}$ was not given. The two values of $x$ that you are given that bracket $e$ are $2$ and $3$, so the correct choice is either $...


$$|x| + |y -x| \ge |x + y -x| = |y|$$ $$|y| + |x -y| \ge |y + x -y| = |x|$$ Move $|x|$ to the right hand side in the first inequality and $|y|$ to the right hand side in the second inequality. We get: $$|y -x| \ge |y| - |x|$$ $$|x -y| \ge |x| -|y|.$$ From absolute value properties we know that $|y-x| = |x-y|$ and if $t \ge a$ and $t \ge −a$ then $t \ge |...


I would write this inequality as $$\frac1{b-a}\int_a^b\exp(g(x))\,dx\ge\exp\left(\frac1{b-a}\int_a^b g(x)\,dx\right).$$ In this guise it is the case of Jensen's inequality for the convex function $\phi(t)=\exp(t)$.


If both $x$ and $y$ are positive then you can just check: $$ \frac{\log(x)}{x} \gt \frac{\log(y)}{y}$$ so if both $x$ and $y$ are greater than $e \approx 2.7183$ then you can just check: $$x \lt y$$


You can simply compute both numbers and compare them: $$ 20! = 2432902008176640000 \\ 2^{40} = 1099511627776 $$ So you can easily see that $2^{40} < 20!$


We have $$\displaystyle \frac{(2n)!}{(n!)^2} = \binom{2n}{n} $$ and of course for every way to choose a combination of $n$ objects from a total of $2n$ objects, there exists a complementary choice (the left over $n$ objects). Since the ways to pick $n$ objects from $2n$ comes in pairs, it follows that the total number is even. Another proof: $$ \frac{(2n)!...


$$2.2^{3.3}\gtrless3.3^{2.2}$$ $$(2.2^3)^{1.1}\gtrless (3.3^2)^{1.1}$$ $$2.2^3\gtrless 3.3^2$$ $$(2\cdot 1.1)^3\gtrless (3\cdot 1.1)^2$$ $$2^3\cdot 1.1\gtrless 3^2$$ $$8\cdot 1.1\gtrless 9$$ $$8.8<9.$$ So, $$2.2^{3.3}<3.3^{2.2}.$$ Can you find now which number is larger: $2.25^{3.375}$ or $3.375^{2.25}$ ;) ?


We could also notice that $3^{43} > 3^{40} = 9^{20} > 8^{20} = 2^{60}$.


Prove $|x| = \max\{x,-x\}$ and $\pm x ≤ |x|$. Then you can use: \begin{align*} a + b &≤ |a| + b ≤ |a| + |b|,\quad\text{and}\\ -a - b &≤ |a| -b ≤ |a| + |b|. \end{align*}


Use polar coordinates: $$x=r \cos \theta,y=r \sin \theta $$ Your inequality becomes $$r^2 \cos \theta \sin \theta \leq r^2 $$ which is pretty much trivial.


Use $e<3$ and $\pi>3$. The first follows from $e:=\lim \left(1+\frac1n\right)^n$ quickly, the second from comparing a circle with its inscribed hexagon.


I don't have any experience with kids, so I have no idea if this would just make things more confusing. But you could try taking advantage of common denominators: Assemble two piles that each contain $15$ identical somethings (paper squares?). Now we can talk about coloring $2/3$ the squares black in the first pile, and coloring $3/5$ of the squares black ...


The first is $7^{91}\times 343$. The second is $7^{91}\times(9/7)^{91}$. Since $\frac{9^3}{7^3}\gt 2$, it follows that $(9/7)^{91}$ is much much bigger than $343$.


In geometry terms that you can understand, the Cauchy-Schwarz inequality says that: Among all the parallelograms with sides a and b, the rectangle is the one with the largest area. Usually you can use this inequality when you are looking for an upper (or lower) bound of an expression. I wanted to give you an example, but my geometry studies are too far ...


By the fundamental theorem of calculus $$e^{i\theta}-1=\int_0^\theta ie^{it}\mathrm{d}t$$ Hence...

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